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Compact Metric Spaces and Inverse Functions ... Apostol, Example After Theorem 4.29 ...

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,891
I am reading Tom M Apostol's book "Mathematical Analysis" (Second Edition) ...

I am focused on Chapter 4: Limits and Continuity ... ...

I need help in order to fully understand the example given after Theorem 4.29 ... ...


Theorem 4.29 (including its proof) and the following example read as follows:




Apostol - 1- Theorem 4.29 & Example  ... PART 1 ...  .png
Apostol - 2 - Theorem 4.29 & Example  ... PART 2 ... .png



In the Example above we read the following:

" ... ... However, \(\displaystyle f^{ -1 }\) is not continuous at the point \(\displaystyle f(0)\). For example, if \(\displaystyle x_n = 1 - 1/n\), the sequence \(\displaystyle \{ f(x_n) \}\) converges to \(\displaystyle f(0)\) but \(\displaystyle \{ x_n \}\) does not converge in \(\displaystyle S\). ... ... "


My question is as follows:

Can someone please explain exactly how/why ... the sequence \(\displaystyle \{ f(x_n) \}\) converges to \(\displaystyle f(0)\) but \(\displaystyle \{ x_n \}\) does not converge in \(\displaystyle S\) ... ... implies that \(\displaystyle f^{ -1 }\) is not continuous at the point \(\displaystyle f(0)\) ... ... ?




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My thoughts ...

I think that the relevant theorem regarding answering my question is Apostol, Theorem 4.16 which reads as follows:


Apostol - Theorem 4.16 ... .png

If we let \(\displaystyle t_n \in f(S)\) be such that \(\displaystyle t_n = f(x_n)\) ... so the sequence \(\displaystyle \{ t_n \} = \{ f(x_n) \}\) is in the domain of \(\displaystyle f^{ -1 }\) ...

Then ... sequence \(\displaystyle \{ t_n \} = \[ f(x_n) \}\) converges to \(\displaystyle f(0) = t_0\) say ...

Then following Theorem 4.16 above ... for \(\displaystyle f^{ -1 }\) to be continuous we need \(\displaystyle \{ f^{ -1 } (t_n) \} = \{ f^{ -1 } ( f(x_n) ) \} = \{ x_n \}\) to converge in \(\displaystyle S\) ... but it does not do so ...

(mind you ... I'm not sure how to prove it doesn't converge in \(\displaystyle S\) ...)

Is that correct?

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Hope that someone can help ...

Peter
 
Last edited:

castor28

Well-known member
MHB Math Scholar
Oct 18, 2017
245
Hi Peter ,

Your idea is correct. Now, the sequence $\{x_n\}$ converges to $1$ in $\mathbb{R}$, but, as $S$ does not contain $1$, that sequence does not converge in $S$.

Intuitively, what happens is that, when a point $P$ travels counterclockwise in circles in $f(S)$, the image $f^{-1}(P)$ jumps discontinuously from $(1-\varepsilon)$ to $0$ in $S$ whenever $P$ passes through $f(0)=1$.
 
Last edited:

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,891
Hi Peter ,

Your idea is correct. Now, the sequence $\{x_n\}$ converges to $1$ in $\mathbb{R}$, but, as $S$ does not contain $1$, that sequence does not converge in $S$.

Intuitively, what happens is that, when a point $P$ travels counterclockwise in circles in $f(S)$, the image $f^{-1}(P)$ jumps discontinuously from $(1-\varepsilon)$ to $0$ in $S$ whenever $P$ passes through $f(0)=1$.




Thanks castor28

Appreciate your help ...

Peter